Let f(14) = 5 and f'(2) > 3 for 14 < I< 20. How small can f(20) be? By the Mean Value Theorem, for some c E (14, 20)
f'(c)
20 - 14
of'(c)(20 - 14)
Of(c)(20 - 14)
f(c)
20 - 14
F(20) - f(14) =
Replacing f(14) by its value, we get f(20)
Since f'(2) >3, we have f(20) >
this the smallest f(20) can be.

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step-by-step explanation:

answer: solve 5\cdot 2^x=2405⋅2  

x

=2405, dot, 2, start superscript, x, end superscript, equals, 240.

solution

to solve for xxx, we must first isolate the exponential part. to do this, divide both sides by 555 as shown below. we do not multiply the 555 and the 222 as this goes against the order of operations!

\begin{aligned} 5\cdot 2^x& =240 2^x& =48 \\ \end{aligned}  

5⋅2  

x

2  

x

​    

=240

=48

​

step-by-step explanation:

the height of the surface increases from the center out to the sides of the road.

step-by-step explanation:

as we move out we move away from the x-axis.

and as mentioned in the problem statement that height of a road increases when we move away from the center x.

hence proved that 3rd option

the height of the surface increases from the center out to the sides of the road.

is the correct option.

step-by-step explanation:

what was your answer?